A 22-1: A circular hoop consists of an 8 lb uniform ring and two point masses, 1

Question

A 22-1: A circular hoop consists of an 8 lb uniform ring and two point masses, 1 lb each, anchored to the ring at the locations shown Starting from rest (i.e. vG 0) and from the position shown, the hoop is given an initial angular velocity of6 rad/s n the counterclockwise direction. The kinetic coefficient of friction between the hoop and the horizontal surface is = 0.3. 6 rad/s 6 in. 30 G 30 (a) Find the angular acceleration as well as the magnitude and direction of the acceleration of G while the hoop rolls and slips on the horizontal surface. (b) Calculate the time taken by the hoop before it stops slipping. Hint: time when vc- (c) Determine the position of P with respect to G at the instant the hoop stops slipping. (d) Calculate the velocity vector of the point mass at P immediately after the hoop stops slipping.

Explanation / Answer

Moment of inertia I = mhoop * r^2 + mr^2 + mr^2

= (8 + 1 + 1) * (6 / 12)^2

I = 2.5 lb-ft^2

a)

Since it is slipping, Friction force f acts towards left. Normal force N acts upwards.

f = mu * N = 0.3 * (8 +1 + 1) = 3 lb

For translatory motion, ma = F + f.......but traction force F = 0

Hence, a = f / m = 3 * 32.2 / (8 + 1 +1) = 9.66 ft/s^2...direction will be towards left.

For rotary motion, Torque T = I*alpha = f*R (clockwise)

alpha = -3 * (6 / 12) * 32.2 / 2.5 = -19.32 rad/s^2

b)

After time t, angular velocity w = w0 + alpha * t

= 6 - 19.32 * t

After time t, translation velocity v = u + at

= 0 + 9.66 * t

Putting v = Rw

9.66 * t = 6 - 19.32 * t

t = 0.207 s

c)

Using theta = wt + 1/2 * alpha * t^2

= 6 * 0.207 - 1/2 * 19.32 * 0.207^2

= 0.828 rad = 47.4 deg

Hence P would have moved by 47.4 degrees. New position of P would be 47.4 - 30 = 17.4 deg

d)

When it stops slipping, v = rw = 9.66 * 0.207 = 2 ft/s

Hence, velocity vector at P would be (-2 i ) + (-2 sin 17.4 i + 2 cos 17.4 j)

= -2.6 i + 1.91 j

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